Totally Ordered Set is Well-Ordered iff Subsets Contain Infima
Theorem
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Then $\struct {S, \preccurlyeq}$ is a well-ordered set if and only if every non-empty subset of $T \subseteq S$ has an infimum such that $\map \inf T \in T$.
Proof
Sufficient Condition
Let $T \subseteq S$ such that $m := \map \inf T \in T$.
By definition, $m$ is a lower bound of $T$ and so:
- $\forall x \in T: m \preccurlyeq x$
That is:
- $\neg \exists x \in T: x \prec m$
Thus by definition $x$ is a minimal element of $T$.
Thus by definition $S$ is well-founded and so is a well-ordered set.
$\Box$
Necessary Condition
Let $\struct {S, \preccurlyeq}$ be a well-ordered set.
Let $T \subseteq S$.
Then from Subset of Well-Ordered Set is Well-Ordered, $\struct {T, \preccurlyeq}$ is also a well-ordered set.
Hence, by definition, $\struct {T, \preccurlyeq}$, has a smallest element $m$.
From Smallest Element is Infimum, $m$ is an infimum.
But $m \in T$.
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings